Integrand size = 35, antiderivative size = 276 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=-\frac {(i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{3/2} d}-\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{3/2} d}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}+\frac {2 (4 A b-3 a B)}{3 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}+\frac {2 b \left (5 a^2 A b+8 A b^3-3 a^3 B-6 a b^2 B\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}} \]
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Time = 1.30 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3690, 3730, 3697, 3696, 95, 209, 212} \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\frac {2 (4 A b-3 a B)}{3 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}+\frac {2 b \left (-3 a^3 B+5 a^2 A b-6 a b^2 B+8 A b^3\right ) \sqrt {\tan (c+d x)}}{3 a^3 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {(-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-b+i a)^{3/2}}-\frac {(B+i A) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (b+i a)^{3/2}}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \]
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Rule 95
Rule 209
Rule 212
Rule 3690
Rule 3696
Rule 3697
Rule 3730
Rubi steps \begin{align*} \text {integral}& = -\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}-\frac {2 \int \frac {\frac {1}{2} (4 A b-3 a B)+\frac {3}{2} a A \tan (c+d x)+2 A b \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx}{3 a} \\ & = -\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}+\frac {2 (4 A b-3 a B)}{3 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}+\frac {4 \int \frac {\frac {1}{4} \left (-3 a^2 A+8 A b^2-6 a b B\right )-\frac {3}{4} a^2 B \tan (c+d x)+\frac {1}{2} b (4 A b-3 a B) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx}{3 a^2} \\ & = -\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}+\frac {2 (4 A b-3 a B)}{3 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}+\frac {2 b \left (5 a^2 A b+8 A b^3-3 a^3 B-6 a b^2 B\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}+\frac {8 \int \frac {-\frac {3}{8} a^3 (a A+b B)+\frac {3}{8} a^3 (A b-a B) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{3 a^3 \left (a^2+b^2\right )} \\ & = -\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}+\frac {2 (4 A b-3 a B)}{3 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}+\frac {2 b \left (5 a^2 A b+8 A b^3-3 a^3 B-6 a b^2 B\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}-\frac {(A-i B) \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 (a-i b)}-\frac {(A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 (a+i b)} \\ & = -\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}+\frac {2 (4 A b-3 a B)}{3 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}+\frac {2 b \left (5 a^2 A b+8 A b^3-3 a^3 B-6 a b^2 B\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}-\frac {(A-i B) \text {Subst}\left (\int \frac {1}{(1-i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a-i b) d}-\frac {(A+i B) \text {Subst}\left (\int \frac {1}{(1+i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a+i b) d} \\ & = -\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}+\frac {2 (4 A b-3 a B)}{3 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}+\frac {2 b \left (5 a^2 A b+8 A b^3-3 a^3 B-6 a b^2 B\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}-\frac {(A-i B) \text {Subst}\left (\int \frac {1}{1-(i a+b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a-i b) d}-\frac {(A+i B) \text {Subst}\left (\int \frac {1}{1-(-i a+b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a+i b) d} \\ & = -\frac {(i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{3/2} d}-\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{3/2} d}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}+\frac {2 (4 A b-3 a B)}{3 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}+\frac {2 b \left (5 a^2 A b+8 A b^3-3 a^3 B-6 a b^2 B\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}} \\ \end{align*}
Time = 3.26 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.08 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\frac {\frac {3 \sqrt [4]{-1} a \left (\frac {(a+i b) (i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {(i a+b) (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}\right )}{a^2+b^2}-\frac {2 A}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}+\frac {8 A b-6 a B}{a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}+\frac {2 b \left (5 a^2 A b+8 A b^3-3 a^3 B-6 a b^2 B\right ) \sqrt {\tan (c+d x)}}{a^2 \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 a d} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 3.99 (sec) , antiderivative size = 9712, normalized size of antiderivative = 35.19
method | result | size |
default | \(\text {Expression too large to display}\) | \(9712\) |
parts | \(\text {Expression too large to display}\) | \(1562265\) |
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Leaf count of result is larger than twice the leaf count of optimal. 18754 vs. \(2 (227) = 454\).
Time = 7.05 (sec) , antiderivative size = 18754, normalized size of antiderivative = 67.95 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]
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\[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {A + B \tan {\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}} \tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]
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Timed out. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}} \,d x \]
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